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chubhunter [2.5K]
3 years ago
7

Alexis and Jessica are shopping. Alexis buys 4 pairs of pants and 3 necklaces and pays $216. Jessica buys 7 pairs of pants and 2

necklaces and pays $300. Solve for the price of each item. Each pair of pants costs _____ dollars Each necklace costs________ dollars
Mathematics
1 answer:
kkurt [141]3 years ago
7 0

Step-by-step explanation:

let us keep the price of pants as <em>x</em>

and price of necklace as <em>y</em>

Simultaneous equations comes as follows:

<em>4x</em><em> </em><em>+</em><em> </em><em>3y</em><em> </em><em>=</em><em> </em><em>2</em><em>1</em><em>6</em><em> </em>for Alexis

<em>7x</em><em> </em><em>+</em><em> </em><em>2y</em><em> </em><em>=</em><em> </em><em>3</em><em>0</em><em>0</em><em> </em><em>for</em><em> </em><em>Jess</em>

<em>we'll</em><em> </em><em>make</em><em> </em><em>either</em><em> </em><em>x</em><em> </em><em>or</em><em> </em><em>y</em><em> </em><em>equal</em><em> </em>

<em>here</em><em> </em><em>let's</em><em> </em><em>make</em><em> </em><em>y</em><em> </em><em>equal</em><em> </em>

<em>2</em><em> </em><em>(</em><em> </em><em>4x</em><em> </em><em>+</em><em> </em><em>3y</em><em> </em><em>=</em><em> </em><em>2</em><em>1</em><em>6</em><em>)</em>

<em>3</em><em> </em><em>(</em><em> </em><em>7x</em><em> </em><em>+</em><em> </em><em>2y</em><em> </em><em>=</em><em> </em><em>3</em><em>0</em><em>0</em><em>)</em>

<em>8x</em><em> </em><em>+</em><em> </em><em>6y</em><em> </em><em>=</em><em> </em><em>4</em><em>3</em><em>2</em>

<em>21x</em><em> </em><em>+</em><em> </em><em>6y</em><em> </em><em>=</em><em> </em><em>9</em><em>0</em><em>0</em>

<em>21x</em><em> </em><em>-</em><em> </em><em>8x</em><em> </em><em>=</em><em> </em><em>9</em><em>0</em><em>0</em><em> </em><em>-</em><em> </em><em>4</em><em>3</em><em>2</em>

<em>13x</em><em> </em><em>=</em><em> </em><em>4</em><em>6</em><em>8</em>

<em>x</em><em> </em><em>=</em><em> </em><em>$</em><em>3</em><em>6</em>

<em>so</em><em> </em><em>a</em><em> </em><em>pant</em><em> </em><em>costs</em><em> </em><em>$</em><em>3</em><em>6</em>

<em>And</em><em> </em>

<em>3y</em><em> </em><em>=</em><em> </em><em>7</em><em>2</em>

<em>y</em><em> </em><em>=</em><em> </em><em>$</em><em> </em><em>2</em><em>4</em>

<em>so</em><em> </em><em>a</em><em> </em><em>necklace</em><em> </em><em>costs</em><em> </em><em>$</em><em>3</em><em>6</em>

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The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
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Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
2 years ago
HELP WILL GIVE BRAINLIEST IF CORRECT
mart [117]

Answer:

B. No, the remainder is -50.

General Formulas and Concepts:

<u>Algebra I</u>

  • Roots are when the polynomial are equal to 0

<u>Algebra II</u>

  • Synthetic Division

Step-by-step explanation:

<u>Step 1: Define</u>

Function f(x) = x³ - 10x² + 27x - 12

Divisor/Root (x + 1)

<u>Step 2: Synthetic Division</u>

<em>See Attachment.</em>

To determine whether a given root is an actual root, the remainder must equal 0. Since we have a remainder of -50, the given root is not a factor of the polynomial.

<em>Please excuse the bad handwriting. Hope this helped!</em>

6 0
3 years ago
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